Question

[Nakamori Aoko]

Chung minh rang:  \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)

Answer 1

[soyeon_Tiểu bàng giải]

Dat A=1/3-2/3²+3/3³-4/3⁴+...+99/3⁹⁹-100/3¹⁰⁰

3A=1-2/3+3/3²-4/3³+...+99/3⁹⁸-100/3⁹⁹

3A+A=1-1/3+1/3²-1/3³+...+1/3⁹⁸-1/3⁹⁹-100/3¹⁰⁰=4A

4A.3=3-1+1/3-1/3²+...+1/3⁹⁷-1/3⁹⁸-100/3⁹⁹=12A

4A+12A=3-100/3⁹⁹-1/3⁹⁹-100/3¹⁰⁰

16A=3-300/3¹⁰⁰-3/3¹⁰⁰-100/3¹⁰⁰=3-403/3¹⁰⁰<3

A<3/16

Chung to...